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zero-one polynomial

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momo
2012-02-13
2012-11-20
  • momo
    momo
    2012-02-13

    Hello,
    I have a problem: all of a polynomial real zero-finding.
    The problem is as follows:
    Be a Real polynomial p (x) = (am) * x ** m + (am-1) * x ** m-1 + …. + a1 * x + a0
    with m = 2 * k and real parameters (Am. .. a0)
    are all zeros of p (x) all real?
    i.e. the output to be rlqe (by quantifier elimination) must be true or false!
    I have tried to solve with realroots and r_solve but no success.
    Thank you very much.

     
  • Thomas Sturm
    Thomas Sturm
    2012-02-13

    I do not quite understand you problem.

    If you have got concrete coefficients, then you can use "realroots" and compare the number of roots found with the degree.

    If you consider a generic problem with all the a_i being parameters, then you can for fixed degree m use real quantifier elimination on

    ex({x1,...,xm}, x1<>x2 and x1<>x3 and ... and x_m-1<>x_m and p(x_1)=0 and ... and p(x_m)=0)
    

    ,

    but for absolutely generic problems you will not get very far.

    Thomas

     
  • momo
    momo
    2012-02-13

    Many thanks for your answer
    my idea is:
    If I have a polynomial of degree m and I m getting zero the real.
    Then the problem is solved.
    Only here is the problem that I work with general real parammetern

     
  • Thomas Sturm
    Thomas Sturm
    2012-02-13

    Quantifier elimination would give you necessary and sufficient conditions in the parameters. Can you post concrete input, and carefully formulate the question that you have on that input?

    Thomas

     
  • momo
    momo
    2012-02-13

    Hallo Thomas,
    Input : something real polynomial with parameters;
    Output : all the real zero the polynomial
    Thanks